Why isn't tangential acceleration just $a$?

If the tangential acceleration is $$\mathrm d|v|/\mathrm dt$$ then isn't it just the magnitude of the acceleration of the object because $$\mathrm dv/\mathrm dt$$ is acceleration?

You've got to be careful with notation when differentiating vectors, because it's very easy to make mistakes.

The acceleration is a vector and is a derivative of the velocity vector $$$$\vec{a} = \frac{d \vec{v}}{d t}$$$$ It is often useful to express this in terms of components. Let's stick with Cartesian coordinates, since then we don't have to worry about derivatives of the basis vectors. If we define $$\hat{e}_i$$ (where $$i=x, y, z$$) to be a set of three unit vectors in the $$x, y, z$$ directions, then since $$d\hat{e}_i/dt$$=0, we have that the components of the acceleration are related to the components of the time derivatives of the velocity vector by $$$$a_i = \hat{e}_i \cdot \vec{a} = \hat{e}_i \cdot \frac{d \vec{v}}{d t} = \frac{d}{dt} \left(\hat{e}_i \cdot \vec{v}\right) = \frac{d v_i}{dt}$$$$ where $$\vec{A}\cdot \vec{B}$$ represents the dot product between $$\vec{A}$$ and $$\vec{B}$$. The magntitude of the acceleration vector is $$$$|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = \sqrt{\vec{a} \cdot \vec{a}}$$$$

Now let us differentiate the magnitude of the velocity vector. The magnitude is given by $$$$|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$$$$ so $$\begin{eqnarray} \frac{d|\vec{v}|}{dt} &=& \frac{1}{2\sqrt{v_x^2 + v_y^2 + v_z^2}} \left(2 v_x \frac{dv_x}{dt} + 2 v_y \frac{dv_y}{dt} + 2 v_z \frac{dv_z}{dt}\right) \\ &=& \frac{v_x a_x + v_y a_y + v_z a_z}{\sqrt{v_x^2 + v_y^2 + v_z^2}} \\ &=& \frac{\vec{v} \cdot \vec{a}}{|\vec{v}|} \end{eqnarray}$$

Comparing the expressions we've derived for $$|\vec{a}|$$ and $$\frac{d|\vec{v}|}{dt}$$, it's clear that they are not equal. For one thing, $$|\vec{a}$$| is always positive, while $$\frac{d|\vec{v}|}{dt}$$ can be positive or negative since $$|\vec{v}|$$ can be growing or shrinking. For another thing, the magnitudes are also different in general. To give a concrete example, if $$a_x=a$$, $$a_y=a_z=0$$, $$v_x=v_y=v/\sqrt{2}$$, and $$v_z=0$$, then $$|\vec{a}|=|a|$$ and $$\frac{d|\vec{v}|}{dt} = a / \sqrt{2}$$.

The acceleration vector has two components.

• The tangential component (tangent to the path) equals the rate of change of speed $$a_t = \dot{v}$$. Speed is the magnitude of the velocity vector $$v = \| \vec{v} \|$$ and the velocity vector defines the tangent direction $$\hat{e} = \vec{v} / \| \vec{v} \\$$

• The normal component (perpendicular to the path) equals the acceleration due to rotation about a point $$a_n = \frac{v^2}{r}$$ where $$r$$ is the radius of curvature.

Together you have

$$\boxed{ \vec{a} = \dot{v} \hat{e} + \frac{v^2}{r} \hat{n} }$$

You can always decompose any acceleration vector into the above components.